Chapter 30: The Nature of the Atom • Rutherford scattering and the nuclear atom –!the nucleus is really " small! • Atomic line spectra – wavelengths characteristic of each element • Bohr model of the hydrogen atom –! " quantization of angular momentum ! quantization of energy • x-rays • The laser • Omit 30.5, 6 – quantum mechanical picture of H-atom, Pauli ! Exclusion Principle, periodic table Monday, April 2, 2007 1 Early model of the atom • About 10-10 m in size. • Positively charged (“pudding”), with negatively charged electrons " (“plums”) embedded in the pudding – “plum pudding model” Stability of the plum pudding was questionable. Where would the characteristic spectral lines come from? # Monday, April 2, 2007 2 Geiger+Marsden: Fall of the plum pudding Alpha (!) particles – nuclei of 4He atom, emitted by some radioactive nuclei – were scattered from a thin gold foil and observed on a screen as flashes of light. Far more were scattered at large angle than would be possible with the weak electric field inside a “plum pudding” atom. Rutherford: realized that the positive charges of the atom had to be contained in a very small volume – the nucleus –!electric " planetary model of the atom with electrons field very strong close to it. " in orbit around nucleus Monday, April 2, 2007 3 Very schematic picture of an atom Size of the nucleus $ 10-15 – 10-14 m Size of the atom $ 10-10 m Equal amounts of + and – charges in the neutral atom. Electrons are in orbit around the nucleus in a planetary model of the atom. (charge +Ze) Monday, April 2, 2007 4 Problem with a Planetary Model of the Atom • The electrons suffer centripetal acceleration in their orbital motion. • Accelerated charges should radiate electromagnetic energy. " The electrons should lose energy and spiral into the nucleus in very " " little time. " A planetary atom should not be stable! " Classical theory does not explain the structure of the atom. " Small systems, such as atoms, must behave differently from ! large. 30.1 Monday, April 2, 2007 5 Fraunhofer absorption lines from the sun http://www.harmsy.freeuk.com/fraunhofer.html 1 Å (Angstrom) = 0.1 nm • Fraunhofer lines – due to absorption of "sunlight " by elements in the atmosphere of the sun • chemical elements emit and absorb light at ! the same wavelengths • Models of the atom need to explain this The sun: blackbody radiation for T $ 6000 K Monday, April 2, 2007 6 Line spectrum of the hydrogen atom n = 2 n = 4 n = 6 n = 3 n = 5 n = 4 m = 1 m = 2 m = 3 n = & Ultraviolet Visible series limit Infrared + other series Balmer found by trial and error a simple formula to calculate the wavelength of all lines of the hydrogen atom: m Series 1 1 1 = R m = 1,2,3... 1 Lyman ultraviolet m2 − n2 ! 2 Balmer visible ! " n = m + 1,m + 2,m + 3... 3 Paschen infrared R = Rydberg constant = 1.097 % 107 m-1 4 Brackett infrared Monday, April 2, 2007 7 Spectrum of H-atom 1 1 1 = R m = 1,2,3... ! m2 − n2 ! " n = m + 1,m + 2,m + 3... R = Rydberg constant = 1.097 % 107 m-1 n = & n = 2 Example: Lyman series, m = 1: For n = &: 1 1 1 m = 1 = 1.097 107 = 1.097 107 m–1 ! × 12 − " × ! " ! = 91.2 nm (series(serieslimit) limit, shortest wavelength in Lyman series) For n = 2: 1 1 1 = 1.097 107 = 0.823 107 m–1 ! × 12 − 22 × ! " ! = 121.5 nm (longest wavelength in Lyman series) Monday, April 2, 2007 8 Bohr model of the hydrogen atom Now superseded by more modern quantum mechanical ideas, but gives the correct answers. E = hf = Ei – Ef Assume: • A planetary model with electron " in orbit around the nucleus. • There are certain electron orbits “Stationary” " that are stable (“stationary states”). states " This is contrary to classical theory " and is not explained in this model. • Light is emitted or absorbed when an electron changes state. • Energy is conserved, so the energy of the photon is the difference in " energy between the initial and final states: " E = hf = Ei – Ef " – what are the allowed energies? Monday, April 2, 2007 9 Energies of hydrogen atom in the Bohr model The electron is kept in orbit by the attractive A hydrogen-like atom or ion with just Coulomb force between it and the nucleus. one electron in orbit kZe2 mv2 F = = r2 r kZe2 mv2 = = 2 KE (A) r The potential energy is: m kZe2 PE = − r And so the total mechanical energy, E = KE + PE, is: kZe2 kZe2 kZe2 E = = (B) 2r − r − 2r But, what are the “good” values of r corresponding to the stationary states? Monday, April 2, 2007 10 Energies of hydrogen atom in the Bohr model Assertion (Bohr): the angular momentum of the electron around the nucleus can have only certain, quantized, values: nh Ln = mvr = n = 1, 2, 3... 2! Combine with (A): mv2r = kZe2 kZe2 nh 2 m So, v2 = = mr 2!mr ! " 2 2 1 nh 11 n r = = (5.29 10− m) n = “quantum number” kZe2m 2! × Z ! " kZe2 And then, substituting into (B), the total mechanical energy is: E = − 2r ! " 2!2mk2e4 Z2 Z2 E = , n = 1,2,3... E = 13.6 eV n − h2 n2 n − n2 ! " Monday, April 2, 2007 11 Bohr Model for H and H-like Atoms • A planetary model in which the electron defies the normal laws of " electromagnetism and does not radiate when in certain “stationary” " states. • The angular momentum of the electron in the stationary states is: nh L = mvr = , n = 1, 2, 3 . n 2π " n is a “quantum number”. Z2 • The energy of the n’th quantum state is: E = 13.6 eV n − n2 • Light is emitted or absorbed when the atom changes energy state. • The energy of the photon is the difference in energy of the two states " of the atom. Monday, April 2, 2007 12 Emission of a photon E = E + h f E = hf i f Absorption of a photon E f + h f = Ei Emission and absorption occur at the same wavelengths (as seen in the E = hf Fraunhofer absorption lines of sunlight). Monday, April 2, 2007 13 Spectrum of hydrogen-like atoms and ions (Just one electron in orbit around the n nucleus) i E = hf = Ei – Ef nf Z2 Energy levels: E = 13.6 eV Z = 1 for hydrogen n − n2 hc 2 1 1 Energy of photons: E = Ei Ef = = 13.6Z 2 2 eV − λ !nf − ni " In agreement with Balmer’s formula 19 (13.6 1.602 10− J) 7 1 R = × × = 1.097 10 m− hc × ! " Monday, April 2, 2007 14 Energy levels of the hydrogen atom – Bohr Free electron # # 13.6 Origin of the lines of the gy gy En = eV − n2 hydrogen atom spectrum Ener Ener (Z = 1) Ground state Monday, April 2, 2007 15 Free electron Prob. 30.-/7: It is possible to use electromagnetic radiation to ionize atoms. To do so, the atoms must absorb the radiation, the photons of which must have enough energy to remove an electron from an atom. What is the longest radiation wavelength # that can be used to ionize the ground state gy of the hydrogen atom? Ener The least energy to remove a ground state electron is E& – E1. Ground state, n = 1 Monday, April 2, 2007 16 Free electron First excited state, n = 2 # gy Prob. 30.7/8: The electron in a hydrogen atom is in the first excited state, when the electron acquires an Ener additional 2.86 eV of energy. What is the quantum number, n, of the state into which the electron moves? Ground state Monday, April 2, 2007 17 Prob. 30.12/14: A hydrogen atom is in the ground state. It absorbs energy and makes a transition to the n = 3 excited state. The atom returns to the ground state emitting two photons. What are their wavelengths? Monday, April 2, 2007 18 Justification of Bohr’s condition on the angular momentum nh Ln = mvr = n = 1, 2, 3... 2! Substitute for the de Broglie wavelength of the electron: h This does NOT show mv = p = the path of the ! electron, but indicates n = 4 hr nh there is a wave moving So, mvr = = around the orbit –!take ! 2" with a pinch of salt! Therefore, 2!r = n" " The circumference of the orbit is an integer number of wavelengths. Condition for a “standing wave” that reinforces itself. Monday, April 2, 2007 19 Bohr’s theory applies to hydrogen-like atoms and ions – that have just one electron in orbit around the nucleus. Examples: H atom, " Z = 1, 1 electron present He+ ion, " Z = 2, 1 electron removed Li2+ ion, " Z = 3, 2 electrons removed Be3+ ion," Z = 4, 3 electrons removed, etc Prob. 30.16: Doubly ionized lithium Li2+ (Z = 3) and triply ionized beryllium Be3+ (Z = 4) each emit a line spectrum. For a certain series of lines in the lithium spectrum, the shortest wavelength if 40.5 nm. For the same series of lines in the beryllium spectrum, what is the shortest wavelength? Both are hydrogen-like atoms, with all electrons removed but one, so Bohr’s formula applies for energies of stationary states: " En = –13.6 Z2/n2 eV Monday, April 2, 2007 20 An x-ray tube Spectrum of x-rays Electrons: KE = eV + Vacuum – • Sharp lines characteristic of the chemical " element of the target. • A smooth background (“Bremsstrahlung”) • A cutoff wavelength, $0, below which there are " no x-rays Monday, April 2, 2007 21 Spectrum of x-rays • Bremsstrahlung: due to the " electrons being stopped " rapidly on hitting the target – " accelerated charges emit " electromagnetic radiation.